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#### Page No 61:

#### Question 1:

A body of mass *m* is placed on a table. The earth is pulling the body with a force *mg*. Taking this force to be the action what is the reaction?

#### Answer:

The Earth is pulling the body with an action force *mg. *This force is exerted by the body on the table. In turn, the table exerts equal and opposite reaction force ($-mg$) on the body in the upwards direction which balances the weight of the body.

#### Page No 61:

#### Question 2:

A boy is sitting on a chair placed on the floor of a room. Write as many action-reaction pairs of forces as you can.

#### Answer:

The given situation involves two action-reaction pairs of forces. They are:

(a) The action force (*F = mg*) with which the boy pushes the chair in the downward direction and the reaction force (*F' = $-$mg*) with which the chair exerts on the boy in the upward direction.

(b) The action force (*F = *(*M + m*)*g*) with which the boy and the chair push the ground in the downward direction and the reaction force (*F' = *(*M + m*)*g*) with which the ground exerts on the boy and the chair in the upward direction.

#### Page No 61:

#### Question 3:

A lawyer alleges in court that the police had forced his client to issue a statement of confession. What kind of force is this?

#### Answer:

The police may beat the lawyer's client until he issues a statement of confession. This force does not have any physical significance in physics. Hence, it is not a force of physics.

#### Page No 61:

#### Question 4:

When you hold a pen and write on your notebook, what kind of force is exerted by you on the pen? By the pen on the notebook? By you on the notebook?

#### Answer:

When we hold a pen and write on our notebook, we are exerting electromagnetic force on the pen. The pen is also exerting the same force on the notebook. We are also exerting gravitational force on the notebook.

#### Page No 61:

#### Question 5:

Is it true that the reaction of a gravitational force is always gravitational, of an electromagnetic force is always electromagnetic and so on?

#### Answer:

Yes, it is true that the reaction of a gravitational force is always gravitational and that of an electromagnetic force is always electromagnetic and so on. For example, if the Earth exerts a gravitational action force on the Moon, then the Moon will also exert the same gravitational reaction force on it. Similarly, when a comb, just after being used, is brought near a piece of paper, it exerts an electromagnetic action force on the paper and the paper will also exert an electromagnetic reaction force on it.

#### Page No 61:

#### Question 6:

Suppose the magnitude of Nuclear force between two protons varies with the distance between them as shown in figure. Estimate the ratio "Nuclear force/Coulomb force" for

(a) *x* = 8 fm

(b) *x* = 4 fm

(c) *x* = 2 fm

(d) *x* = 1 fm (1 fm = 10 ^{−15}m).

Figure

#### Answer:

First let us calculate the coulomb force between 2 protons for distance = 8 fm

$F\mathit{}\mathit{=}\frac{\mathit{K}\mathit{}{\mathit{q}}^{\mathit{2}}}{{\mathit{r}}^{\mathit{2}}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times (1.6\times {10}^{-19}{)}^{2}}{(8\times {10}^{-15}{)}^{2}}\phantom{\rule{0ex}{0ex}}=3.6\mathrm{N}$

${F}_{\mathrm{N}}=0.05\mathrm{N}\phantom{\rule{0ex}{0ex}}\frac{{F}_{\mathrm{N}}}{{F}_{\mathrm{C}}}=\frac{0.05}{3.6}=0.0138\mathrm{N}$

For x= 4 fm

${F}_{\mathrm{C}}=\frac{9\times {10}^{9}\times (1.6\times {10}^{-19}{)}^{2}}{(4\times {10}^{-15}{)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{23.04\times {10}^{-29}}{(4\times {10}^{-15}{)}^{2}}\phantom{\rule{0ex}{0ex}}=14.4\mathrm{N}\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{N}}=1\mathrm{N}\phantom{\rule{0ex}{0ex}}\frac{{F}_{\mathrm{N}}}{{F}_{\mathrm{C}}}=\frac{1}{14.4}=0.0694\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{For}x=2\mathrm{fm}\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{C}}=\frac{9\times {10}^{9}\times (1.6\times {10}^{-19}{)}^{2}}{(2\times {10}^{-15}{)}^{2}}\phantom{\rule{0ex}{0ex}}=57.6\mathrm{N}\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{N}}=10\mathrm{N}\phantom{\rule{0ex}{0ex}}\frac{{F}_{\mathrm{N}}}{{F}_{\mathrm{C}}}=\frac{10}{57.6}=0.173\phantom{\rule{0ex}{0ex}}\mathrm{For}x=1\mathrm{fm}\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{C}}=\frac{9\times {10}^{9}\times (1.6\times {10}^{-19}{)}^{2}}{(1\times {10}^{-15}{)}^{2}}\phantom{\rule{0ex}{0ex}}=230.4\mathrm{N}\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{N}}=1000\mathrm{N}\phantom{\rule{0ex}{0ex}}\frac{{F}_{\mathrm{N}}}{{F}_{\mathrm{C}}}=\frac{1000}{230.4}=4.34$

#### Page No 61:

#### Question 7:

List all the forces acting on the block B in figure.

Figure

#### Answer:

Forces acting on block B:

- A pair of electromagnetic forces between the blocks A and B
*.* - A pair of electromagnetic forces between the blocks B and the Earth.
- A pair of gravitational forces between the blocks B and the Earth.
- A pair of gravitational forces between the blocks A and B.
- A pair of action-reaction forces between the man and block A
*.* - Weight of the man, block A and block B as action forces on the ground in the downward direction.
- Reaction of the ground on the man, block A and block B in the upward direction to balance their weights.

#### Page No 61:

#### Question 8:

List all the forces acting on (a) the pulley *A*, (b) the boy and (c) the block *C* in figure.

Figure

#### Answer:

**Forces acting on the pulley A:**

- A tension exerts electromagnetic force between the string and pulley A.
- A pair of gravitational force between the Earth and pulley.

**Forces acting on the boy:**

- A tension exerts electromagnetic force between the string and the boy.
- A pair of gravitational force between the Earth and the boy.

**Forces acting on the block C**:

- A tension exerts electromagnetic force between the string and block
*.* - A pair gravitational force between block C and the Earth.

#### Page No 61:

#### Question 9:

Figure shows a boy pulling a wagon on a road. List as many forces as you can which are relevant with this figure. Find the pairs of forces connected by Newton's third law of motion.

Figure

#### Answer:

List of forces:

(a) A pair of gravitational force between the wagon and the Earth.

(b) A frictional force exerted by the road on the wagon.

(c) A tension exerts electromagnetic force between the wagon and string.

(d) A pair of gravitational force between the man and the Earth.

(e) A frictional force exerted by the road on the man.

(f) A tension exerts electromagnetic force between the man and string.

Here, (a), (c), (d) and (f) are pairs of forces associated with Newton's third law of motion.

#### Page No 61:

#### Question 10:

Figure shows a cart. Complete the table shown below.

Figure

Force on | Force by | Nature of the force | Direction |

Cart |
1 2 3 : |
||

Horse |
1 2 3 : |
||

Driver |
1 2 3 : |

#### Answer:

Force on | Force by | Nature of the force | Direction |

Cart |
1. Horse 2. Road 3. Earth |
1. Action force 2. Electromagnetic force exerts frictional force. 3. Gravitational Force |
1. Forward 2. Forward 3. Downward |

Horse |
1. Cart 2. Road 3. Road 3. Earth |
1. Reaction Force 2. Electromagnetic force exerts frictional force 3. Reaction force 4. Gravitational force |
1. Backward 2. Forward 3. Forward 4. Downward |

Driver |
1. Cart 2. Rope 3. Earth |
1. Electromagnetic force 2. Tension exerts electromagnetic force 3. Gravitational force |
1. Backward 2. Forward 3. Downward |

#### Page No 62:

#### Question 1:

When Neils Bohr shook hand with Werner Heisenberg, what kind of force they exerted?

(a) Gravitational

(b) Electromagnetic

(c) Nuclear

(d) Weak.

#### Answer:

(b) Electromagnetic

When Neils Bohr shook hand with Werner Heisenberg, electromagnetic force is exerted on each other. This is because when their hands made contact with each other, the atoms at the two surfaces come close to each other. The charged constituents of the atoms in the hands exert forces on each other and, as a result, a measurable force is produced.

#### Page No 62:

#### Question 2:

Let E, G and N represent the magnitudes of electromagnetic gravitational and nuclear forces between two electrons at a given separation. Then

(a) N>E>G

(c) E>N>G

(c) G>N>E

(d) E>G>N

#### Answer:

(d) E>G>N

Between two electrons at a given separation, the strongest acting force is the electromagnetic force. The gravitational force is the weakest force between any two particles. There is no nuclear force acting between them, because it exists only in the nucleus (between proton-proton or neutron-neutron or both).

#### Page No 62:

#### Question 3:

The sum of all electromagnetic forces between different particles of a system of charged particles is zero

(a) only if all the particles are positively charged

(b) only if all the particles are negatively charged

(c) only if half the particles are positively charged and half are negatively charged

(d) irrespective of the signs of the charges.

#### Answer:

(d) irrespective of the signs of the charges.

The sum of all electromagnetic forces between different particles of a system of charged particles is zero irrespective of the sign of the charges, because electromagnetic force is a vector quantity that depends upon the direction. So, we consider the directions while adding vector quantities.

#### Page No 62:

#### Question 4:

A 60 kg man pushes a 40 kg man by a force of 60 N. The 40 kg man has pushed the other man with a force of

(a) 40 N

(b) 0 N

(c) 60 N

(d) 20 N

#### Answer:

(c) 60 N

According to Newton's third law, which states that an action-reaction pair of forces are equal in magnitude, the man who weighs 40 kg will push the other man with the same force of 60 N.

#### Page No 62:

#### Question 1:

A neutron exerts a force on a proton which is

(a) gravitational

(b) electromagnetic

(c) nuclear

(d) weak

#### Answer:

(a) gravitational

(c) nuclear

A neutron exerts both the gravitational and nuclear forces on a proton.

The gravitational force can be seen between a neutron and a proton. However, its strength is negligible. The nuclear force is exerted only if the interacting particles are neutrons or protons or both. A neutron cannot exert electromagnetic force, because it is a neutral particle.

#### Page No 62:

#### Question 2:

A proton exerts a force on a proton which is

(a) gravitational

(b) electromagnetic

(c) nuclear

(d) weak

#### Answer:

(a) gravitational

(b) electromagnetic

(c) nuclear

A proton exerts gravitational, electromagnetic and nuclear forces on a proton.

The gravitational force can easily be seen everywhere; it the weakest among all the forces. It is negligible in elementary particles like protons.

The electromagnetic force can be seen between charged particles. Since proton is a charged particle, it can exert this force on other protons.

The nuclear force is present only if the interacting particles are protons or neutrons or both. So, this force can exist between two protons.

#### Page No 62:

#### Question 3:

Mark the correct statements:

(a) The nuclear force between two protons is always greater than the electromagnetic force between them.

(b) The electromagnetic force between two protons is always greater than the gravitational force between them.

(c) The gravitational force between two protons may be greater than the nuclear force between them.

(d) Electromagnetic force between two protons may be greater than the nuclear force acting between them.

#### Answer:

(b) The electromagnetic force between two protons is always greater than the gravitational force between them.

(c) The gravitational force between two protons may be greater than the nuclear force between them.

(d) Electromagnetic force between two protons may be greater than the nuclear force acting between them.

The electromagnetic force between two protons is always greater than the gravitational force between them, because gravitational force is the weakest force in nature.

We know that nuclear force is the strongest force in nature when the distance between two particles is less than ${10}^{-14}\mathrm{m}$. However, the gravitational as well as electromagnetic forces between two protons may be greater than the nuclear force acting between them when the distance between them is more than ${10}^{-14}\mathrm{m}$.

#### Page No 62:

#### Question 4:

If all matters were made of electrically neutral particles such as neutrons,

(a) there would be no force of friction

(b) there would be no tension in the string

(c) it would not be possible to sit on a chair

(d) the earth could not move around the sun.

#### Answer:

(a) there would be no force of friction

(b) there would be no tension in the string

(c) it would not be possible to sit on a chair

For the existence of friction between two bodies and tension in a string, electromagnetic force is needed. Electromagnetic force exists only between charged particles. For sitting on a chair, we need fractional force. A neutral particle can exert gravitational force on other neutral particles. So, even if all the matters were made up of electrically neutral particles, the earth will still move around the sun.

#### Page No 62:

#### Question 5:

Which of the following systems may be adequately described by classical physics?

(a) motion of a cricket ball

(b) motion of a dust particle

(c) a hydrogen atom

(d) a neutron changing to a proton.

#### Answer:

(a) motion of a cricket ball

(b) motion of a dust particle

Classical physics adequately describes the motion of a ball and the motion of dust particles. It also describes Newton's law of motion, Newton's law of gravitation, Laws of thermodynamics, Maxwell's electromagnetism and Lorentz force. Classical physics easily describes the system of heavenly bodies like the Sun, the Earth and the Moon. However, it is inadequate for describing systems of particles which have sizes much smaller than ${10}^{-6}\mathrm{m}$ (e.g., atom, nuclei and other elementary particles).

#### Page No 62:

#### Question 6:

The two ends of a spring are displaced along the length of the spring. All displacement have equal magnitudes. In which case or cases the tension or compression in the spring will have a maximum magnitude?

(a) the right end is displaced towards right and the left end towards left

(b) both ends are displaced towards right

(c) both ends are displaced towards left

(d) the right end is displaced towards left and the left end towards right.

#### Answer:

(a) the right end is displaced towards right and the left end towards left

(d) the right end is displaced towards left and the left end towards right.

When the right end is displaced towards the right and the left end towards the left, then this is the case of tension (expansion) and the spring will have maximum displacement.

When the right end is displaced towards the left and the left end towards the right, then this is the case of compression and ,again, the spring will have maximum displacement.

#### Page No 62:

#### Question 7:

Action and reaction

(a) act on two different objects

(b) have equal magnitude

(c) have opposite directions

(d) have resultant zero.

#### Answer:

(a) act on two different objects

(b) have equal magnitude

(c) have opposite directions

(d) have resultant zero.

The two forces $\overrightarrow{F}\mathrm{and}-\overrightarrow{F}$ connected by Newton's third law are known as action-reaction pair.

For example, when a man jumps out from a ferry, he applied some force on the ferry. So, the action force in this case is the force applied by the man and the reaction is the force exerted by the ferry on the man. Both the forces act upon different objects (the man and the ferry) but have equal magnitudes and opposite directions. As a result, their resultant is zero.

#### Page No 63:

#### Question 1:

The gravitational force acting on a particle of 1 g due to a similar particle is equal to 6.67 × 10^{−17} N. Calculate the separation between the particles.

#### Answer:

Mass of the particle, *m *= 1 gm$=\frac{1}{1000}\mathrm{kg}$

Let the distance between the two particles be *r.*

Gravitational force between the particle, *F* = 6.67 × 10^{−17} N

Also, $\mathrm{F}=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}$

Substituting the respective values in the above formula, we get:

$6.67\times {10}^{-17}=\frac{6.67\times {10}^{-11}\times \left(1/1000\right)\times \left(1/1000\right)}{{r}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{6.67\times {10}^{-6}\times {10}^{-11}}{6.67\times {10}^{-17}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-17}}{{10}^{-17}}=1\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{1}=1\mathrm{m}$

∴ The separation between the particles is 1 m.

#### Page No 63:

#### Question 2:

Calculate the force with which you attract the earth.

#### Answer:

Consider that a man is standing on the surface of the Earth.

Force acting on the man = *mg*

Here, *m* = mass of the man and *g* = acceleration due to gravity on the surface of earth (=10 m/s^{2})

Assume that the mass of the man is equal to 65 kg.

Then *F* = *W* = *mg* = 65 × 10 = 650 N = force acting on the man

∴ By Newton's third law (action-reaction are always equal), the man is also attracting the earth with a force of 650 N in the opposite direction.

#### Page No 63:

#### Question 3:

At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?

#### Answer:

Given: ${q}_{1}={q}_{2}=1\mathrm{C}$

By Coulomb's law, the force of attraction between the two charges is given by

$F=\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times 1\times 1}{{r}^{2}}$

However, the force of attraction is equal to the weight (*F = mg*).

$\therefore mg=\frac{9\times {10}^{9}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{9\times {10}^{9}}{m\times 10}=\frac{9\times {10}^{8}}{m}(\mathrm{Taking}g=10\mathrm{m}/{\mathrm{s}}^{2})\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{9\times {10}^{8}}{m}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{3\times {10}^{4}}{\sqrt{m}}$

Assuming that *m* = 81 kg, we have:

$r=\frac{3\times {10}^{4}}{\sqrt{81}}\phantom{\rule{0ex}{0ex}}=\frac{3}{9}\times {10}^{4}\mathrm{m}\phantom{\rule{0ex}{0ex}}=3333.3\mathrm{m}$

∴ The distance *r* is 3333.3 m.

#### Page No 63:

#### Question 4:

Two spherical bodies, each of mass 50 kg, are placed at a separation of 20 cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.

#### Answer:

Mass = 50 kg

Separation between the masses, *r* = 20 cm = 0.2 m

Let the change on each sphere be *q*.

Now, gravitational force, ${F}_{\mathrm{G}}=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}$

$=\frac{67\times {10}^{-11}\times {\left(50\right)}^{2}}{{\left(0.2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{67\times {10}^{-11}\times 2500}{0.04}$

$\mathrm{Coulomb}\mathrm{force},{F}_{\mathrm{c}}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=9\times {10}^{9}\frac{{q}^{2}}{0.04}$

Since *F*_{G} = *F*_{c}, we have:

$\frac{6.7\times {10}^{-11}\times 2500}{0.04}=\frac{9\times {10}^{9}\times {q}^{2}}{0.04}\phantom{\rule{0ex}{0ex}}\Rightarrow {q}^{2}=\frac{6.7\times {10}^{-11}\times 2500}{9\times {10}^{9}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}}{0.04}=1809\times {10}^{-18}\phantom{\rule{0ex}{0ex}}\therefore q=\sqrt{18.09\times {10}^{-18}}\phantom{\rule{0ex}{0ex}}=4.3\times {10}^{-9}\mathrm{C}$

Thus, the charge of the spherical body is $4.3\times {10}^{-9}\mathrm{C}$.

#### Page No 63:

#### Question 5:

A monkey is sitting on a tree limb. The limb exerts a normal force of 48 N and a frictional force of 20 N. Find the magnitude of the total force exerted by the limb on the monkey.

#### Answer:

Given: The limb of the tree exerts a normal force of 48 N and a frictional force of 20 N.

So, resultant magnitude of the force if given by

$R=\sqrt{\left({48}^{2}+{20}^{2}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{2304+400}\phantom{\rule{0ex}{0ex}}=\sqrt{2704}=52\mathrm{N}$

∴ The magnitude of the total force exerted by the limb on the monkey is 52 N.

#### Page No 63:

#### Question 6:

A body builder exerts a force of 150 N against a bullworker and compresses it by 20 cm. Calculate the spring constant of the spring in the bullworker.

#### Answer:

Force exerted by the body builder against the bullworker = 150 N

Compression in the bullworker, *x* = 20 cm = 0.2 m

∴ Total force exerted, *f* = *kx* = 150

Here, *k *is the spring constant of the spring in the bullworker.

$\therefore k=\frac{150}{0.2}=\frac{1500}{2}=750\mathrm{N}/\mathrm{m}$

Hence, the spring constant of the spring in the bullworker is 750 N/m.

#### Page No 63:

#### Question 7:

A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.)

#### Answer:

Let *h *be the height, *M *be the Earth's mass, *R *be the Earth's radius and *m *be the satellite's mass

.

Force on the satellite due to the earth when it is at the Earth's surface, ${F}_{1}=\frac{\mathrm{GM}m}{{\mathrm{R}}^{2}}$

Force on the satellite due to the earth when it is at height *h* above the Earth's surface, ${F}_{2}=\frac{\mathrm{GM}m}{{\left(\mathrm{R}\mathit{+}h\right)}^{2}}$

According to question, we have:

$\frac{{F}_{1}}{{F}_{2}}=\frac{{\left(R+h\right)}^{2}}{{R}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2=\frac{{\left(R+h\right)}^{2}}{{R}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{squareroot}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\sqrt{2}=1+\frac{h}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow h=\left(\sqrt{2}-1\right)R\phantom{\rule{0ex}{0ex}}=0.414\times 6400=2649.6\mathrm{km}\approx 2650\mathrm{km}$

#### Page No 63:

#### Question 8:

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force of the separation is increased to 25 cm?

#### Answer:

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other.

So, ${F}_{1}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\xb7\frac{{q}^{2}}{{r}_{1}^{2}}$

Also, ${F}_{2}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\xb7\frac{{q}^{2}}{{r}_{2}^{2}}$

According to the question, we have:

$\frac{{F}_{2}}{{F}_{1}}=\frac{{r}_{1}^{2}}{{r}_{2}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{20\times 20}{25\times 25}=\frac{16}{25}\phantom{\rule{0ex}{0ex}}\therefore {F}_{2}=\frac{16}{25}\times {F}_{1}$_{}

Therefore, the two charged particles will exert a force of 13.0 N on each other, if the separation is increased to 25 cm.

#### Page No 63:

#### Question 9:

The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data : The universal constant of gravitation G = 6.67 × 11^{−11} N−m^{2}/kg^{2}, mass of the moon = 7.36 × 10^{22} kg, mass of the earth = 6 × 10^{24} kg and the distance between the earth and the moon = 3.8 × 10^{5} km.

#### Answer:

The force between the Earth and the Moon is given by $F=\frac{\mathrm{G}Mm}{{r}^{2}}$.

Here*, M *is the mass of the earth; *m* is the mass of the moon and *r* is the distance between Earth and Moon.

On substituting the values, we get:

$F=\frac{6.67\times {10}^{-11}\times 7.36\times {10}^{22}\times 6\times {10}^{24}}{3.8\times 3.8\times {10}^{16}}$

$=\frac{6.67\times 7.36\times {10}^{35}}{(3.8{)}^{2}\times {10}^{16}}\phantom{\rule{0ex}{0ex}}=20.3\times {10}^{19}=2.03\times {10}^{20}\phantom{\rule{0ex}{0ex}}\approx 2.0\times {10}^{20}\mathrm{N}\phantom{\rule{0ex}{0ex}}$

∴ The weight of the moon is $2.0\times {10}^{20}\mathrm{N}$.

#### Page No 63:

#### Question 10:

Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

#### Answer:

Charge of the proton, *q* = $1.6\times {10}^{-19}\mathrm{C}$

Mass of the proton = $1.67\times {10}^{-27}\mathrm{kg}$

Let the distance between two protons be *r. *

Coulomb force (electric force) between the protons is given by

${f}_{e}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\times \frac{{q}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times (1.6{)}^{2}\times {10}^{-38}}{{r}^{2}}$

Gravitational force between the protons is given by

${f}_{g}=\frac{\mathrm{G}{m}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{6.67\times {10}^{-11}\times (1.67\times {10}^{-27}{)}^{2}}{{r}^{2}}$

On dividing ${f}_{e}\mathrm{by}{f}_{g}$, we get:

$\frac{{f}_{e}}{{f}_{g}}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\times \frac{{q}^{2}}{{r}^{2}}\times \frac{{r}^{2}}{\mathrm{G}{m}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times 1.6\times 1.6\times {10}^{-38}}{6.67\times {10}^{-11}\times 1.67\times 1.67\times {10}^{-54}}\phantom{\rule{0ex}{0ex}}=\frac{9\times (1.6{)}^{2}\times {10}^{-29}}{6.67\times (1.67{)}^{2}\times {10}^{-65}}\phantom{\rule{0ex}{0ex}}=1.24\times {10}^{36}$

#### Page No 63:

#### Question 11:

The average separation between the proton and the electron in a hydrogen atom in ground state is 5.3 × 10^{−11} m. (a) Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?

#### Answer:

Average separation between the proton and the electron of a Hydrogen atom in ground state, *r* = 5.3 × 10^{−11} m

(a) Coulomb force when the proton and the electron in a hydrogen atom in ground state

$F=9\times {10}^{9}\times \frac{{q}_{1}{q}_{2}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{\left(5.3\times {10}^{-11}\right)}^{2}}=8.2\times {10}^{-8}\mathrm{N}$

(b) Coulomb force when the average distance between the proton and the electron becomes 4 times that of its ground state

$\mathrm{Coulomb}\mathrm{force},F=\frac{1}{4\mathrm{\pi}{\in}_{0}}=\frac{{q}_{1}{q}_{2}}{{\left(4r\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{16\times {\left(5.3\right)}^{2}\times {10}^{-22}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {\left(1.6\right)}^{2}}{10\times {\left(5.3\right)}^{2}}\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=0.0512\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=5.1\times {10}^{-9}\mathrm{N}$

#### Page No 63:

#### Question 12:

The geostationary orbit of the earth is at a distance of about 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.

#### Answer:

The geostationary orbit of the Earth is at a distance of about 36000 km.

We know that the value acceleration due to gravity above the surface of the Earth is given by $g\text{'}=\frac{\mathrm{G}m}{{\left(R+h\right)}^{2}}$.

At *h* = 36000 km, we have:

$g\text{'}=\frac{\mathrm{G}m}{{\left(36000+6400\right)}^{2}}$

At the surface, we have:

$g=\frac{\mathrm{G}m}{{\left(6400\right)}^{2}}\phantom{\rule{0ex}{0ex}}\therefore \frac{g\text{'}}{g}=\frac{6400\times 6400}{42400\times 42400}\phantom{\rule{0ex}{0ex}}=\frac{256}{106\times 106}=0.0228\phantom{\rule{0ex}{0ex}}\Rightarrow g\text{'}=0.0227\times 9.8=0.223\left[\mathrm{Taking}g=9.8\mathrm{m}/{\mathrm{s}}^{2}\mathrm{at}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{earth}\right]$

For a 120 kg equipment placed in a geostationary satellite, its weight will be

*mg*' = 120 × 0.233

$\Rightarrow $26.76 ≈ 27 N

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